Algebra II



Algebra II

Segment 1 EXAM STUDY GUIDE

← Variations (Module 1):

o Direct variation: y = kx where k is the constant

o Inverse or indirect variation: y = k/x

o Joint variation: y = kxz

← Slope & Equations of lines (Module 1):

o Given two points on a line, use the slope formula to find the slope: (y2 – y1) / (x2 – x1)

▪ Find the slope of (8, -2) and (-2, 1): (1 - -2) / (-2 – 8) = 3/-10

o Given the equation of a line, write the line in slope-intercept form (y = mx +b) and identify the slope (m).

▪ Example 3x + 4y = 8 ( subtract 3x to get 4y = -3x + 8 (

divide by 4 on each term y = -3/4x + 2 ( Then the slope is -3/4 and the y- ` intercept of the line is (0,2)

o Point-slope form: y – y1 = m(x – x1) where m is the slope and (x1, y1) is a point on the line

▪ Example: Given (3,2) and m = -1/3 write an equation of the line

y – 2 = -1/3 (x – 3) (distribute the -1/3 inside the parenthesis)

y – 2 = -1/3x – 1 (Add 2 to both sides of the equation)

y = -1/3x + 1 This is slope-intercept form

1/3x + y = 1 (Multiply each term by 3)

1x + 3y = 3 This is standard form

← Finding x-and y-intercepts (Module 1):

o When finding the x-intercept, let y = 0 and solve for x.

▪ Example: 4x + 12y = 48

4x + 12(0) = 48

4x = 48

X = 12 x-intercept is then (12,0)

o When finding the y-intercept, let x = 0 and solve for y.

▪ Example: 4x + 12y = 48

4(0) + 12y = 48

12y = 48

y = 4 y-intercept is then (0, 4)

← Horizontal and Vertical lines (Module 1):

o Horizontal lines have slopes of zero and are of the form y = ___ a number

o Vertical lines have undefined slopes and are of the form x = ___ a number

▪ Example: Through the point (3, -4), the vertical line is x = 3 and the horizontal line is y = -4

← Parallel and perpendicular lines (Module 1):

o Parallel lines have THE SAME SLOPE

▪ If a line has a slope of ¾, a parallel line also has a slope of ¾

o Perpendicular lines have OPPOSITE RECIPROCAL slopes

▪ If a line has a slope of ¾, a perpendicular line then has a slope of -4/3

← Absolute value equations and inequalities (Module 1):

o EQUATIONS: Split apart into TWO equations to solve, one positive and one negative

▪ Example: |4x – 7| = 9

4x – 7 = 9 or 4x – 7 = -9

Now you can solve each one to find the answer

- Remember that you must isolate the absolute value FIRST before separating

▪ Example: |x + 9| - 4 = 10 Add 4 FIRST

|x+9| = 14 THEN split apart into a positive and a negative

X + 9 = 14 or x + 9 = -14

o INEQUALITIES: You split apart into TWO inequalities, leave the first sign the same. For the second inequality, you FLIP the sign AND make the number negative

- Use “OR” for > or ≥; use “AND” for < and ≤

▪ Example: |x + 6| > 10

X + 6 > 10 OR x + 6 < -10

- Finding the VERTEX of an inequality

▪ Standard form is y < (x – x1) + y1 then the vertex is (x1, y1)

← Solving systems of equations (Module 2):

o Graphing: Rearrange the equations into slope-intercepts method and graph, where the two lines intersect will be the solution

▪ Parallel lines will have NO SOLUTION, they will have same slopes but different y-intercepts

▪ Same lines will have ALL REAL solutions or INFINITELY MANY solutions, they will have same slopes AND same y-intercepts

o Substitution: Solve one of the equations for a single variable and then replace that variable in the second equation

o Elimination: Make a pair of opposite coefficients that will be able to cancel out.

o The solution to system of equations is an ordered pair

← Graphing Inequalities in two variables/solving systems of inequalities (Module 2):

o < and > are dotted lines, ≤ and ≥ are solid lines

o < or ≤ means to shade below the line

o > or ≥ means to shade above the line

← Polynomials (Module 3):

o Degree: Add up the exponents for each term, whichever number is the highest is the degree of the polynomial

o Order: Ascending vs. Descending

o Adding: Combine like terms, variables stay the same

▪ Example: 3x + 4y + 2x –8y = 5x – 4y

o Subtracting: Distribute the negative to the parenthesis and then combine like terms

▪ Example: (-2a-7b)-(8a-2b) = -2a – 7b – 8a + 2b = -10a – 5b

o Multiplying: Multiply coefficients, add exponents

▪ Example: 4x3 * 6x5 = 24x8

o Dividing: Divide coefficients, subtract exponents; negative coefficients get moved to the denominator and become positive.

▪ Example: 14x7y4 / 7x5y9 = 2x2y-5 = 2x2/y5

o Multiplying Binomials: Use FOIL (first, outer, inner, last) to multiply each term of the first binomial by each term of the second binomial.

▪ Example: (5x-3)(2x+4)= 5x*2x + 5x*4 + -3*2x + -3*4

= 10x2 + 20x – 6x – 12 = 10x2 + 14x – 12

← Special Products (Module 3):

o Difference of squares: a2 – b2 = (a + b)(a – b)

▪ Example: 49x4 – 9 = (7x2 – 3) (7x2 + 3)

o Perfect square trinomial: a2 – 2ab + b2 = (a – b)2 OR a2 + 2ab + b2 = (a + b)2

▪ Example: 36x2 – 24x + 4 = (6x – 2)2

← Factoring (Module 3):

o Greatest Common Factor – take out the GCF of each term and put it in front of the parenthesis

o Special Products – look for one of the two special products above

o Factor by grouping – four terms, factor the GCF of the first two and last two terms, look to see if you can create two binomials from this

o Create a fourth term to do factoring by grouping

← Simplifying Radicals (Module 4):

o Make sure you identify the index when you’re simplifying the radicals

o Combine only like radicals when adding or subtracting

o When multiplying radicals, multiply the numbers under the radicand and then simplify

o When simplifying radicals, you can change the radical to an expression with a rational exponent (fraction) in order to make simplifying simpler

o Conjugates are similar but have different signs in between terms. For example, a + bi and a – bi are conjugates. When you multiply a binomial by its conjugate, the “i” terms will cancel. You can also do this if you have a radical in your binomial.

o Cycles of i – To simplify an “i” with an exponent higher than 4, divide the exponent by 4 and find the remainder. The cycle for i is i1 = i, i2 = -1, i3 = - i, i4 = 1

← Standard of a Quadratic Equation (Module 5):

o Standard form of a quadratic is y = a(x – h)2 + k

o The “a” tells you which way the quadratic will open, if “a” is positive the quadratic opens UP, if “a” is negative the quadratics opens DOWN.

o The vertex of the quadratic is (h, k)

o You can complete the square in order to get a quadratic equation into standard form

← Solving quadratic equations (Module 5):

o You can factor quadratics, use the quadratic formula if necessary

o Set the factors equal to zero and solve for x to find the solutions (zeros or roots of the quadratic equation)

o Use b2 – 4ac (the discriminant) to see how many times the quadratic will cross the x-axis

▪ If the discriminant is positive, the graph will cross the x-axis TWICE and there will be two real roots

▪ If the discriminant is zero, the graph will cross the x-axis ONCE and there will be one real root

▪ If the discriminant is negative, the graph will NOT cross the x-axis and there will be no real roots (two imaginary roots)

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